ELECTROCHEMISTRY
MCQS
Q.1 Electrolysis is the process in which a chemical reaction takes place at the expense of
(a) chemical energy (b) electrical energy
(c) heat energy (d) none of these
Q.2 Standard hydrogen electrode has an arbitrarily fixed potential
(a) 0.00 volt (b) 1.00 volt
(c) 0.10 volt (d) none of these
Q.3 The oxidation number of chromium in K2Cr2O7 is
(a) 14 (b) 12
(c) 6 (d) none of these
Q.4 In the reaction 2 Fe + Cl2 ( 2FeCl3
(a) Fe is reduced (b) Fe is oxidized
(c) Cl2 is oxidized (d) none of these
Q.5 When fused PbBr2 is electrolyzed
(a) bromine appears at cathode
(b) lead is deposited at the cathode
(c) lead appears at the anode
(d) none of these happens
Q.6 When aqueous solution of NaCl is electrolysed
(a) Cl2 is evolved at the cathode
(b) H2 is evolved at cathode
(c) Na is deposited at the cathode
(d) Na appears at the anode
Q.7 During electrolysis of KNO3, H2 is evolved at
(a) anode (b) cathode
(c) both (a) and (b) (d) none of these
Q.8 During electrolysis of CuSO4 (aq) using Cu electrodes Cu is deposited at
(a) anode (b) cathode
(c) both (a) and (b) (d) none of these
Q.9 During electrolysis of fused NaCl, which of the following reaction occurs at anode
(a) Cl– ions oxidized (b) Cl– ions reduced
(c) Na+ ions oxidized (d) Na+ ions reduced
Q.10 An electrochemical cell is based upon
(a) acid–base reaction (b) redox reaction
(c) nuclear reaction (d) none of the above
Q.11 Which one of the following will be good conductor of electricity
(a) pure distilled water (b) molten NaCl
(c) dilute solution of glucose
(d) chloroform
Q.12 Which one of the following represents the same net reaction as the electrolysis of aqueous H2SO4
(a) electrolysis of water
(b) electrolysis of molten NaCl
(c) electrolysis of aqueous HCl
(d) electrolysis of aqueous NaCl
Q.13 In a galvanic cell, the reaction occurs
2H2O ( O2 (g) + 4H+ + 4e– It occurs at the
(a) cathode (b) anode
(c) cathode and anode (d) none of the above
Q.14 Which statement below is not true for the reaction
Fe3+ + e– ( Fe2+
(a) Fe3+ is reduced
(b) oxidation state of Fe has changed
(c) Fe3+ can act as an oxidizing agent
(d) both Fe2+ and Fe3+ are called anions
Q.15 During a redox reaction, an oxidizing agent
(a) gains electrons (b) is oxidized
(c) loses electrons (d) is hydrolysed
Q.16 In a salt bridge KCl is used because
(a) it is an electrolyte
(b) K+ and Cl– transfer easily
(c) agar–agar forms a good jelly with it
(d) KCl is also present in the calomel electrode
Q.17 A oxidizing agent is a substance which brings about
(a) electron donation (b) oxidation
(c) reduction (d) hydrolysis
Q.18 In the electrolysis the process of oxidation occurs at
(a) anode (b) cathode
(c) both cathode and anode
(d) in electrolytic solution
Q.19 In an oxidation process the oxidation number of the element
(a) increases (b) decreases
(c) does not change (d)
Q.20 In the reduction process the oxidation number of the element
(a) increases (b) decreases
(c) does not change (d)
Q.21 Oxidation number of oxygen in OF2 is
(a) + 1 (b) – 1
(c) + 2 (d) – 2
Q.22 The e.m.f. of Zn – Cu cell is
(a) 1.10 v (b) 1.5 v
(c) 2.0 v (d) 2.5 v
Q.23 The standard reduction potential of a standard hydrogen electrode
(a) 0.0 v (b) 1.1 v
(c) 1.5 v (d) 2.0 v
Q.24 The oxidation number of Mn is K2 MnO4 is
(a) + 2 (b) + 4
(c) + 6 (d) + 7
Q.25 Which of the following is the definition of oxidation
(a) gain of electrons (b) loss of electrons
(c) addition of H2 (d) removal of O2
Q.26 During electrolysis of H2SO4 (aq) O2 is evolved at
(a) cathode (b) anode
(c) both a and b (d) none of these
Q.27 The e.m.f. produced by a voltage cell is
(a) electrode potential (b) reduction potential
(c) cell potential (d) oxidation potential
Q.28 Which of the following is not a redox reaction
(a) CaCO3 ( CaO + CO2
(b) Cu + 4HNO3 ( Cu(NO3)2 + 2NO2 + H2O
(c) 2H2 + O2 ( 2H2O
(d) MnO2 + 4HCl ( MnCl2 + Cl2 + 2H2O
Q.29 Which element acts as a reducing agent in the reaction
Zn + H2SO4 ( ZnSO4 + H2
(a) Zn (b) H
(c) S (d) O
Q.30 Which element acts as a oxidizing agent in the reaction
MnO2 + 4HCl ( MnCl2 + Cl2 + 2H2O
(a) Mn (b) O
(c) H (d) Cl
Q.31 When the current is passed through an electrolytic solution, which of the following process will occur
(a) anions move towards anode and cations move towards cathode
(b) cations and anions both move towards anode
(c) cations and anions both move towards anode
(d) no movement of the ions occur
Q.32 Electric current passes through both molten and solution form of NaCl because of
(a) ionic bonding (b) Na+ and Cl– ions
(c) ions of water (d) hydration of ions
Q.33 A cell which produces electric current by redox reaction is called
(a) standard cell (b) voltaic cell
(c) reversible cell (d) concentration cell
Q.34 Which of the following conduct electricity due to the migration of electrons only
(a) copper metal (b) NaCl molten
(c) NaCl (d) NaCl solution
Q.35 Oxidation number of sulphur in S2O eq \a\co1(2–,3 ) is
(a) + 6 (b) – 2
(c) + 2 (d) + 4
Q.36 Substances through which electric current can pass are called
(a) insulators (b) conductors
(c) cathode (d) anode
Q.37 Substances through which electric current cannot pass are called
(a) insulators (b) conductors
(c) anode (d) cathode
Q.38 Metallic conduction is due to the
(a) movement of electrons
(b) movement of ions
(c) both (a) and (b)
(d) none of these
Q.39 Metallic conductors conduct electricity
(a) with chemical change
(b) without any chemical change
(c) both (a) and (b)
(d) none of these
Q.40 The flow of electrons is called
(a) electrolyte (b) electric current
(c) cathode (d) anode
Q.41 A substance which in molten state or in solution form allows electric current to pass through it is called
(a) electrolyte (b) insulator
(c) conduction (d) none of these
Q.42 The process in which electric current is used to carry out a non–spontaneous redox reaction is called
(a) electrolyte (b) electrolysis
(c) metallic conductor (d) electrodes
Q.43 In electrochemical cells, the electrode at which the reduction occurs is called
(a) anode (b) cathode
(c) electrolyte (d) electrolysis
Q.44 The process of producing a chemical change in an electrolytic cell is called
(a) electrolyte (b) electrolysis
(c) electrodes (d) conductor
Q.45 The process in which ionic compound when fused or dissolved in water split up into charged particles is called
(a) electrolysis (b) hydration
(c) ionization (d) conduction
Q.46 An apparatus in which chemical energy in converted to electrical energy is called
(a) electrolytic cell (b) galvanic cell
(c) fuel cell (d) down cell
Q.47 The metallic conductors in contact with the solution are called
(a) insulator (b) electrodes
(c) electrolyte (d) down cell
Q.48 The reaction in a galvanic cell is
(a) spontaneous (b) non–spontaneous
(c) acid–base (d) none of these
Q.49 Caustic soda is obtained by electrolysis of conc. aqueous solution of NaCl in a cell called
(a) Daniell’s cell (b) Nelson’s cell
(c) Down’s cell (d) Voltaic cell
Q.50 Sodium metal is obtained by the electrolysis of fused NaCl in a cell is called
(a) Nelson’s cell (b) Down’s cell
(c) Daniell cell (d) Voltaic cell
Q.51 The e.m.f. of Daniell cell can be increased by
(a) increasing the area of electrode
(b) increasing the concentration of oxidising ion in the solution
(c) increasing the concentration of reducing ion in the solution
(d) adding the dil H2SO4
Q.52 Metal and their ionic salts both conduct electricity. Which of the following statement is not correct both
(a) are good conductors normally
(b) are ionic in nature
(c) decompose on passing current
(d) are normally solid
Q.53 The branch of chemistry which deals with the relationships between electricity and chemical reaction is called
(a) chemical kinetics (b) electrochemistry
(c) stiochiometry (d) thermochemistry
Q.54 A system containing of electrodes that dips into an electrolyte in which a chemical reaction either uses or generates an electric current is called
(a) voltaic cell (b) electrochemical cell
(c) voltaic or galvanic cell (d) fuel cell
Q.55 A cell in which spontaneous redox reaction generates an electric current is called
(a) electrolytic cell
(b) electrochemical cell
(c) voltaic orgalvanic cell
(d) biological cell
Q.56 A cell in which an electric current drives a non–spontaneous reaction is called
(a) electrolytic cell (b) voltaic cell
(c) biological cell (d) electrochemical cell
Q.57 A process for converting one metal with a thin layer of another metal is called
(a) electrolysis (b) electroplating
(c) electrode potential (d) standard electrode
Q.58 In an electrical connection between cathode and anode of a voltaic cell, electrons flow from the
(a) anode to the cathode (b) cathode to the anode
(c) both (a) and (b) (d) none of these
Q.59 Greater the value of standard reduction potential of a species indicates
(a) greater its tendency to accepted electrons
(b) lesser tendency to accept electrons
(c) greater tendency to lose electrons
(d) none of these
Q.60 In lead accumulator the electrolyte H2SO4 solution is
(a) 30 % (b) 60% H2SO4
(c) 80% (d) 90%
Q.61 In alkaline battery, the electrolyte contains
(a) MnO2 (b) KOH
(c) NaCl (d) NaNO3
Q.62 Alkali metals have
(a) lower value of reduction potential than coinage metals
(b) higher value of reduction potential than coinage metals
(c) equal values of reduction potential to coinage metals
(d) none of these
Q.63 Strong reducing agents have
(a) greater positive value of standard reduction potential
(b) greater negative value of standard reduction potential
(c) lesser positive value of standard reduction potential
(d) none of these
Q.64 Strong oxidizing agents have
(a) greater positive value of standard reduction potential
(b) lesser positive value of standard reduction potential
(c) greater negative value of standard reduction potential
(d) none of these
Q.65 The electrode with more negative value of reduction potential acts as
(a) cathode (b) anode
(c) electrode (d) none of these
Q.66 Metals which are above SHE in electrochemical series
(a) can liberate H2 from acid
(b) cannot liberate H2 from acid
(c) cannot always liberate H2 from acid
(c) none of these
Q.67 Corrosion reactions are
(a) spontaneous redox reactions
(b) non–spontaneous redox reactions
(c) spontaneous acid–base reactions
(d) none of these
Q.68 Voltaic cell can be changed into
(a) electrochemical cell (b) electrolytic cell
(c) reversible cell (d) primary cell
Q.69 Strongest oxidizing agent in the electrochemical series is
(a) Li (b) F
(c) H2 (d) I2
Q.70 Strongest reducing agent in the electrochemical series is
(a) Li (b) F
(c) H2 (d) I2
Q.71 Fuel cells are the means by which chemical energy may be converted into
(a) heat energy (b) electrical energy
(c) mechanical energy (d) sound energy
ANSWERS
Questions 1 2 3 4 5 Answers b a c b b Questions 6 7 8 9 10 Answers b b b a b Questions 11 12 13 14 15 Answers b a b d a Questions 16 17 18 19 20 Answers b c a a b Questions 21 22 23 24 25 Answers c a a c b Questions 26 27 28 29 30 Answers b c a a a Questions 31 32 33 34 35 Answers a b b a c Questions 36 37 38 39 40 Answers b a a b b Questions 41 42 43 44 45 Answers a b b b c Questions 46 47 48 49 50 Answers b b a b b Questions 51 52 53 54 55 Answers c b b b c Questions 56 57 58 59 60 Answers a b a a a Questions 61 62 63 64 65 Answers b a b a b Questions 66 67 68 69 70 Answers a a c b a Questions 71 Answers b
SHORT QUESTION WITH ANSWERS
Q.1 Differentiate electrolytic and voltaic cell.
Ans.
An electrochemical cell in which electric current is used to drive a non–spontaneous redox reaction is called electrolytic cell. e.g. electrolysis of molten NaCl.
An electrochemical cell in which electricity is produced as a result of spontaneous redox reaction is called Galvanic or voltaic cell e.g. Zn – Cu cell.
Q.2 Differentiate b/w ionization and electrolysis.
Ans.
Ionization Electrolysis 1. The process of splitting up of an ionic compound into charged particles when fused or dissolved in water is called ionization. 1. The process of decomposing a substance usually in solution or in molten state is called electrolysis. 2. In this process, there is no need of electrodes. 2. In this process, electrodes are required. 3. There is no need of electricity for ionization. 3. There is a need of electricity for electrolysis. 4. During ionization ions are produced. 4. In electrolysis, ions are oxidized or reduced to neutral atoms or molecules. Q.3 What are conductors? Define electronic conductors and electrolytic conductors.
Ans. Conductor:
The substance through which electricity can pass is called conductor. There are two types of conductors.
Electronic Conductor:
The conductors in which electricity is passed due to the movement of free electrons e.g. metals.
Electrolytic Conductors:
The conductors in which electricity is passed, due to movement of ions to their respective electrodes.
Q.4 Differentiate b/w oxidation and reduction.
Ans. Oxidation:
(i) Addition of oxygen
(ii) Removal of electrons or
(iii) Removal of hydrogen is called as oxidation.
Reduction:
(i) Addition of hydrogen
(ii) Addition of electrons or
(iii) Removal of oxygen is called as reduction.
Q.5 Define oxidation number. Write down the rules for assigning oxidation number.
Ans. Oxidation Number:
The apparent charge present on an atom in a molecule or ionic compound is called oxidation number.
Rules:
(1) Oxidation state of all elements in free state is zero e.g.
O eq \a\co1(0,2) , N eq \a\co1(0,2) , F eq \a\co1(0,2) , Cl eq \a\co1(0,2) etc.
(2) Oxidation number of hydrogen is + 1 except in case of metal hydride where its is – 1 H+1Cl and NaH–1.
(3) Oxidation number of oxygen is – 2 except in case of per oxides and super oxide. Where it is – 1 and – eq \f(1,2) respectively.
(4) In binary compounds of halogens, the charge of halogen is
– 1 e.g. NaCl–1.
(5) Oxidation number of IA group elements in combined form is + 1, Group II is + 2 and group III is + 3.
(6) The sum of oxidation state of all elements in neutral compounds is zero e.g. K+1M+7nO eq \a\co1(–2(4),4 ) .
(7) The sum of oxidation states of ion is equal to charge present on it P+5O eq \a\co1(–3,4 ) eq \x( P = + 5 )
Q.6 K2Cr2O7 + HCl ( KCl + CrCl3 + Cl2 + H2O
Calculate the oxidation number of each element in above equation?
Ans.
K eq \a\co1(+1(2),2 ) Cr eq \a\co1(+6(2),2 ) O eq \a\co1(–2(7),7 ) + H+1Cl–1 (
K+1Cl–1 + Cr+3 Cl eq \a\co1(–1(3),3 ) + Cl eq \a\co1(0,2) + H eq \a\co1(+1(2),2 ) O
Q.7 Calculate oxidation number of Cr in the following compounds.
(i) CrCl3 (ii) Cr2(SO4)3 (iii) K2CrO4
(iv) K2Cr2O7 (v) CrO3 (vi) Cr2O3
(vii) Cr2O eq \a\co1(2–,7 )
Ans.
(i) CrCl3 ( x + (–1) 3 = 0 x = 0 + 3
(ii) Cr2(SO4)3 ( 2x + 6 (3) + (–2) 4 (3) = 0
= 2x + 18 – 24 = 0
2x – 6 = 0
2x = 6 eq \x( x = 3 )
(iii) K2CrO4 ( + 1 (2) + x + – 2 (4) = 0
= 2 + x – 8 = 0 eq \x( x = + 6 )
(iv) K2Cr2O7 ( + 1 (2) + 2x + – 2 (7) = 0
+ 2 + 2x – 14 = 0
2x – 12 = 0
2x = 12
eq \x( x = + 6 )
(v) CrO3 ( x + (–2) 3 = 0
eq \x( x = + 6 )
(vi) Cr2O3 ( 2x + (–2) 3 = 0
2x – 6 = 0
2x = + 6
eq \x(x = + 3 )
(vii) Cr2O eq \a\co1(–2,7 ) ( 2x + (–2) 7 = – 2
2x – 14 = – 2
2x = + 12
eq \x( x = + 6 )
Q.8 In a Galvanic cell chemical energy is converted to electrical energy. How?
Ans.
In galvanic cell chemical energy changes to electrical energy because in this cell redox reaction takes place spontaneously. As a result of which electrons are transferred from anode to cathode. These electron flow is called current or electrical energy.
Q.9 Zn acts as anode when connected to Cu but as cathode when connected to Al?
Ans.
Zn acts as anode when connected to Cu because reduction potential of Cu is greater than Zn. So Zn will be oxidised and acts as anode Cu acts as cathode. But in case of Al connected to Zn. Zinc acts strong oxidising agent (with greater reduction potential) than Al so Al will act as anode and Zn acts as cathode.
Q.10 Explain through equations how lead battery is discharged.
Ans.
In this process, the anode and cathode of an external source are joined to the anode and cathode of the cell respectively. Then the redox cathode of the cell respectively. Then the redox reactions at respective electrodes are reversed. Hence cell begins to recharge reaction at cathode:
Pb eq \a\co1(+2 ,(aq)) + Ze– ( Pb eq \a\co1(0 ,(s)) (Reduction)
PbSO4(s)
SO eq \a\co1(2–,4 ) (aq)
Reaction at Anode:
+2H2O
Pb+2(aq) ( PbO2 + OH+ + 2e–
PbSO4(s)
SO eq \a\co1(2– ,4(aq))
Overall Reaction:
2PbSO4(s) + 2H2O(l) ( Pb(s) + PbO2(s) + 4H+ + 2SO eq \a\co1(2– ,4(aq))
Q.11 What is density of acid used in lead accumulator and voltage of cell?
Ans.
When lead accumulator is in use it acid conc. falls and density decrease to 1.15 g/cm3. Its e.m.f. also decreases. After recharging conc. of acid is increased to make up its density to 1.25 gcm–3 and voltage becomes 2 volts.
Q.12 What are advantages of Alkaline battery?
Ans.
(1) It can be used over a wider range of temp due to more stability of electrolysis.
(2) It requires very little electrolyte so it is very compact.
(3) It maintains a constant voltage for a longer period of time and therefore lasts longer.
Q.13 What are silver oxide batteries?
Ans.
These are tiny and expensive batteries, which are commonly used in electronic watches, auto exposure camera and calculators. These are small button shape cells. In these types of batteries silver oxide (Ag2O) mixed with NaOH or KOH acts as cathode and zinc functions as anode.
Q.14 Discuss reactions taking place in the Ni–Cd cell.
Ans.
It is strong cell. The anode is made up of cadmium at which oxidation reaction takes place while cathode consists of NiO2. In this cell, an alkaline electrolyte is used.
Reactions:
At anode:
Cd(s) + 2OH eq \a\co1(– ,(aq)) ( Cd(OH)2(s) + 2e–
NiO2(s) + 2H2O(l) + 2e– ( Ni(OH)2(s) + 20H–
Overall Reactions:
Cd(s) + NiO2(s) + 2H2O ( Cd(OH)2(s) + Ni(OH)2(s)
Q.15 How fuel is converted into electrical energy in fuel cells?
Ans.
A type of cell, which is similar to galvanic cell in which, fuel used may be gaseous or liquid substance e.g. hydrogen, hydrazine, methanol, ammonia etc. and oxidants are generally oxygen or air. Fuel and oxidant are supplied to respective electrodes, which are porous and activated with catalyst.
Reactions:
At anode:
H2(g) + 2OH eq \a\co1(– ,(aq)) ( 2H2O(l) + 2e– (oxidation)
At cathode:
O2(g) + 2H2O(l) + 4e– ( 4OH eq \a\co1(– ,(aq)) (Reduction)
Overall reaction
H2(g) + O2(g) ( 2H2O(l)
Q.16 What is importance of fuel cells?
Ans.
The fuel cells are of great importance and used in space vehicles. In these cells, electrodes are made up of porous compressed carbon, which is impregnated with platinum. Which acts as catalyst. This fuel cell works at high temp in order to evaporate water, which is formed in it. After condensation, this water can be used for drinking water for astronauts.
Q.17 How prediction about feasibility of chemical reaction is made by electrochemical series?
Ans.
For a particular reaction, it is easy to predict whether it will take place or not. This is done by using electrochemical series e.g. Cu+2 can oxidise zinc metal but Zn+2 cannot oxidise Cu metal because Cu is below Zn in the electrochemical series. i.e.
Cu eq \a\co1(+2 ,(aq)) + Zn(s) ( Cu(s) + Zn eq \a\co1(2+,(aq)) . E eq \a\co1(0 ,cell) = 1.10 v
Cu(s) + Zn+2 ( Cu+2 + Zn(s) E eq \a\co1(0 ,cell) = – 1 . 1
This is not feasible.
Q.18 How e.m.f. of galvanic cell is calculated with the help of electrochemical series?
Ans.
The standard electrode potential values can be used to determine e.m.f. of a given galvanic cell i.e. e.m.f. of Daniel cell is calculated as
Zn(s) | Zn eq \a\co1(+2,(aq)) || Cu eq \a\co1(2+,(aq)) | Cu(s)
(Anode) Cathode
(Oxidation half cell) Reduction half cell)
Reactions:
At anode
Zn ( Zn2+ + 2e– E0 = + 0.76 v
+2e–
Cu+2 ( Cu(s) E0 = + 0.34 v
E eq \a\co1(0 ,cell) = E eq \a\co1(0 ,Anode) + E eq \a\co1(0 ,Cathode)
= 0.76 + 0.34 = 1.10 v
Q.19 What are relative reactivity of metals? Explain by electrochemical series?
Ans.
Reactivity of a metal depends upon its ability to lose electrons to change into M+ cation. It is clear from electrochemical series that smaller the value of standard reduction potential, greater its tendency to lose electrons. Alkali metals (Li, NaK) having lower value of reduction potentials are highly reactive then coinage metals
(Cu, Ag, Au).
Q.20 How it is possible to select a element as anode or cathode?
Ans.
The electrode, which has lower value of reduction potential acts as anode and that which has higher value of reduction potential acts as cathode E0 eq \b( \f(Cu+2,Cu) ) = 0.34 has a lower value of reduction potential than eq \f(E\a\co1(0 ,Ag+),Ag) = 0.80 v anode and Ag as cathode.
Q.21 How we can predict whether it takes place or not?
Ans.
With the help of electrochemical series, it can be determined whether a reaction will occur spontaneously or not. Consider the following reaction
Pb+2 + 2Ag0 ( 2Ag+ + Pb0
The reaction will occur only if E eq \a\co1(0 ,Cell) is positive
E eq \a\co1(0 ,Cell) = E eq \a\co1(0 ,Anode) + E eq \a\co1(0 ,Cathode)
E eq \a\co1(0 ,Cell) = E eq \a\co1(0 ,2Ag/Ag+) + E eq \a\co1(0 ,Pb2+/Pb)
E eq \a\co1(0 ,Cell) = – 0.80 + (– 0.13)
= – 0.93
E eq \a\co1(0 ,Cell) is negative so reaction will not take place spontaneously.
Q.22 How can a non–metal displace another non–metal?
Ans.
A non–metal of higher standard reduction potential can displace other non–metal of lower standard reduction potential
Cl2 + MgBr2 ( MgCl2 + Br2
or Cl2(g) + 2Br eq \a\co1(– ,(aq)) ( 2Cl eq \a\co1(– ,(aq)) + Br2(g)
In above reaction E eq \a\co1(0 ,Cl2/2Cl–) = + 1.36 v
is greater than E eq \a\co1(0 ,Br2/2Br–) = + 1.06 v.
Q.23 What is standard hydrogen electrode (SHE)?
Ans.
Standard hydrogen electrode is used to determine the electrode potential of other electrodes. It is used as reference electrode and its value is taken as 0.0 v.
SHE consist of platinum foil, coated with finally divided black platinum and suspended in 1 M HCl solution. Pure H2 gas at one atm pressure is bubbled into 1M HCl solution. The value of this half cell will be zero either it is oxidised or reduced
2H+ + 2e– ( H2 0.00 V (Reduction)
H2 ( 2H+ + 2e– 0.00 V (Oxidation)
Q.24 What is difference b/w cell and battery?
Ans.
The arrangement in which electrical energy is converted to chemical energy or chemical energy is converted to electrical energy is called cell.
The combination of two or more cells is called battery.
Q.25 Outline the important applications of electrolysis?
Ans.
(1) Electroplating
(2) Electro–refining
(3) Electro–manufacturing
(4) Electrotyping
Q.26 What is salt bridge? Also give its function?
Ans:
It is U-shaped glass tube having saturated solution of strong electrolyte like KCl. The glass tube is filled with jelly type material
It has two major functions:
It connects the two solutions in two half cells
It maintains the electrical neutrality by the diffusion of ions through it.
Q.27 What is the difference between primary and secondary cell?
Ans:
The cell which is not rechargeable called primary cell e.g Dry cell
The cell which is rechargeable called secondary cell. e.g lead accumulator
Q.28 What is electrochemical series?
Ans:
A list of elements in which they are arranged in the order of their standard electrode potential on hydrogen scale is called electrochemical series.
TEXT BOOK EXERCISE
Q1. Multiple choice questions.
(i) The cathode reaction in the electrolysis of dill. H2SO4 with Pt electrode is
(a) Reduction (b) Oxidation
(c) Both oxidation and reduction
(d) Neither oxidation nor reduction
(ii) Which of the following statement is correct about galvanic cell?
(a) Anode is negatively charged
(b) Reduction occurs at anode
(c) Cathode is positively charged
(d) Reduction occurs at cathode
(iii) Stronger the oxidizing agent, greater is the
(a) Oxidation potential
(b) Reduction potential
(c) Redox potential
(d) EMF of cell
(iv) If the slat bridge is not used between two half cells, then the voltage
(a) Decrease rapidly (b) Decrease slowly
(c) Does not change (d) Drops to zero
(v) If a strip of Cu metal is placed in a solution of FeSO4.
(a) Cu will be precipitated out
(b) Fe is precipitated out
(c) Cu and Fe both dissolve
(d) No reaction takes place
Ans. i) a ii) d iii) b iv) d v) d
Q2. Fill in the blanks with suitable words.
(i) The oxidation number of O atom is _____in OF2 and is _____in H2O2.
(ii) Conductivity of metallic conductors is due to the flow of _____while that of electrolytes is due to flow of ______.
(iii) Reaction taking place at the ______is termed as oxidation, and at the _____ is called reduction.
(iv) _____is setup when a metal is dipped in its own ions.
(v) Cu metal _____the Cu-cathode when electrolysis is performed for CuSO4 solution with Cu-cathodes.
(vi) The reduction potential of Zn is _____volts and its oxidation potential is ______volts.
(vii) In a fuel cell ______react together in the presence of ______.
Ans. i) +2, -1 ii) electrons. Ions
iii) anode, cathode iv) Equilibrium
v) deposits on vi) -0.76, +0.76
vii) H2 and O2, Pt catalyst
Q3. Indicate TURE or FALSE as the case may be.
(i) In electrolytic conduction electrons flow through the electrolyte.
(ii) In the process of electrolysis, the electrons in the external circuit flow from cathode to anode.
(iii) Sugar is a non-electrolyte in solid form and when dissolved in water will allow the passage of an electric current.
(iv) A metal will only allow the passage of an electric current when it is in cold state.
(v) The electrolyte products of aqueous copper (II) chloride solution are copper and chlorine.
(vi) Zinc can displace iron form its solution.
(vii) SHE cats as cathode when connected with Cu electrode.
(viii) A voltaic cell produces electrical energy at the expense of chemical energy.
(ix) Lead storage battery is not reversible cell.
(x) Cr changes its oxidation number when K2 Cr2 O7 is reacted with HCI.
Ans. i) False ii) False iii) False iv) False
v) True vi) True vii) False viii) True
ix) False x) True
Q4. Describe the electrolysis of molten sodium chloride, and a concentrated solution of sodium chloride.
See Section 10.2.4
Q5. What is the difference between single electrode potential and standard electrode potential? How can it be measured? Give its importance.
Q6. Outline the important applications of electrolysis. Also write the electrochemical reactions involved therein. Discuss the electrolysis of Cu SO4 using Cu electrode and AgNO3 solutions using Ag electrode.
See Section 10.2.2
Q7. Describe the construction and working of standard hydrogen electrode.
See Section 10.3.1
Q8. Is the reaction Fe3+ + Ag EMBED Equation.DSMT4 Fe2+ + Ag+ spontaneous? If not, write spontaneous reaction involving these species.
Solution
Fe3+ + Ago EMBED Equation.DSMT4 Fe2+ + Ag+
In this reaction, Fe is reduced while Ag is oxidized. Therefore, Fe+3 will act as cathode while Ago as anode.
Thus, emf of the cell will be
Eocell =Eoox + Eored
Eocell = - 0.7994 + (- 0.44)
Eocell = - 0.7994 - 0.44
Eocell = - 1.2394
Since emf of cell is negative, therefore, the cell-reaction is non-spontaneous.
But if the electrodes are reversed, the cell-reaction becomes spontaneous i.e.
Fe3+ + Ago EMBED Equation.DSMT4 Fe3+ + Ago
Q9. Explain the difference between
(a) Ionization and electrolysis
IONIZATION ELECTROLYSIS 1 The process in which ionic compounds when fused or dissolved in water split up into charged particles called ions. 1 The process in which electricity is used to carry out a non-spontaneous reaction is called electrolysis. 2 Electrodes are not needed 2 Electrodes are required 3 Electricity is not needed 3 Electricity is required 4 Since there are no electrodes, therefore, ions do not move towards electrodes 4 Ions moves towards their respective electrodes 5 After ionization, ions are not discharged 5 Ions are discharged at electrodes to give neutral products.
(b) Electrolytic and Voltaic cell
See Section 10.2.2 and 10.2.5
(c) Conduction through metals and molten electrolytes
CONDUCTION
THROUGH METALS CONDUCTION THROUGH
MOLTEN ELECTROLYTE 1 Electrical conduction takes place due to free electrons. 1 Electrical conduction takes place due to ions 2 There in no need to convert metal into molten state. 2 Electrolyte must be converted into molten state for electrical conduction 3 In this case, conductance decreases with increase in temperature. 3 In this case, conductance increase with increase in temperature. 4 No chemical reaction occurs during conduction. 4 Chemical reaction occurs take place during conduction. 5 Chemical composition of metal is not changed during conduction and no new substance are produced. 5 Since chemical reactions occur, therefore new substances are produced. 6 Example:
All metals are conductions. e.g. Fe, Pb etc. 6 Example:
Molten salt e.g. NaCI(l) or their aqueous solutions, acids, bases etc.
Q10. Describe a galvanic cell explaining the function of electrodes and salt bridge.
See Section 10.2.5
Q11. Write comprehensive notes on
Spontaneity of oxidation-reduction reactions.
See Section 10.4.1
(b) Electrolytic conduction
See Section 10.2
(c) Alkaline, silver oxide and Nickle-Cadmium batteries, fuel cells.
See Section 10.5
(d) Lead accumulator, its desirable and undesirable features.
See Section 10.5.1
Q12. Will the reaction be spontaneous for the following set of half reactions? What will be the value of Ecell?
(i) Cr3+(aq) + 3e- EMBED Equation.DSMT4 Cr(s)
(ii) MnO2(s) + 4H+ + 2e- EMBED Equation.DSMT4 Mn2+(aq) + 2H2O(l)
Standard reduction potential for reaction
(i) = - 0.7 and for the reaction (ii) = + 1.28 V.
In reaction (i) Cr is reduced from +3 to +2.
In reaction (ii) Mn is also reduction reactions; hence these reactions are not possible in these forms.
However, if reaction (i) is reversed so that Cr is oxidized then the reaction becomes spontaneous and its emf can be calculated as
(i) Cr(s) EMBED Equation.DSMT4 Cr3+(aq) + 3e- Eoox = + 0.74V
(ii) MnO2(s) + 4H+ + 2e- EMBED Equation.DSMT4 Mn2+(aq) + 2H2O(l) Eo(aq) =+1.28 V
emf of the cell is given by
Eocell =Eooxl + Eored
Eocel = +0.74 + 1.28
Eocell =2.02 V
Q13. Explain the following with reasons
A porous plate or a salt bridge is not required in lead storage cell.
A salt bridge has two main functions
It joins solution of two half cells and thus complete the circuit.
It maintains electrical neutrality of the two half cells as ions can pass through it.
In lead storage battery, both cathode and anode are dipped in the same solution. Therefore, excess positive or negative ions are not produced in the solution. Hence, there is no need of salt bridge.
The standard oxidation potential of Zn is 0.76 V and its reduction potential is – 0.76 V.
According to the law of conservation of energy, energy can neither be created nor destroyed. Therefore, if standard oxidation potential of Zn is 0.76 V, then its potential for reverse process, i.e. standard reduction potential will also be same but with positive sign. Thus
Zn EMBED Equation.DSMT4 Zn2+ + 2e- Eoox =0.76 V
Zn2+ + 2e- EMBED Equation.DSMT4 Zn Eoox =0.76 V
(C) Na and K can displace hydrogen from acids but Pt, Pd and Cu cannot.
Greater the value of reduction potential, Lesser is the ability to loose electron to from positive ion, Hence weaker is its tendency to displace H2-.
Metals like Pt, Pb, and Cu have high positive value of reduction potential. Thus these do not liberate H2.
Metals like Na and K have negative values of reduction potential. Thus, these can liberate H2.
2Na +2HCI EMBED Equation.DSMT4 2NaCI + H2
2K +2HCI EMBED Equation.DSMT4 2KCI + H2
The equilibrium is set up between metal atoms of electrode and ions of metal in a cell.
When a metal electrode is dipped into the solution of its own ion. there may be two tendencies
Metal atom from electrode leaves the electron on metal an goes into solution. this is oxidation process
M EMBED Equation.DSMT4 M+ + e-
Metal ions in solution may take up electrons form the metal electrode and deposit as atom on electrode. This is reduction process.
M+ e- EMBED Equation.DSMT4 M+
At last, a dynamic equilibrium is established due to same rate of two processes. Thus no further potential difference is developed.
(e) A salt bridge maintains the electrical neutrality in the cell.
Two half cells are electrically connected by a salt bridge.
Consider a Zn-Cu cell
During reactions of this cell, Zn half cell continuously loose electrons.
Thus, in this positive charge is increasing.
Zn EMBED Equation.DSMT4 Zn2+ + 2e-
While, Cu half cell continuously receive electrons, thus it goes on collecting negative charge.
Cu2+ + 2e- EMBED Equation.DSMT4 Cu
Collection of positive charge in Zn electrode half cell and collection of negative charge in Cu half cell would stop the reaction.
Salt bridge prevents the net accumulation of charges in either beaker. Thus form negative Cu half cell, negative ions diffuse through the salt bridge into the positive Zn half cell. In this way, salt bridge maintains the two solution, electrically neutral.
(f) lead accumulator is a chargeable battery.
See Section 10.5.1
(g) Impure Cu can be purified by electrolytic process.
Impure Cu can be made pure in an electrolytic cell. Thick sheets of impure copper are made anode, while, thin sheets of pure copper are made cathode in the cell. These sheets are placed in an electrolytic solution of CuSO4.
When current is passed through the cell, Cu from anode is oxidized to Cu2+ ions, which go into the solution. from the solution, Cu2+ ions are reduced to metallic Cu and deposits as pure Cu on cathode. In this way, impure sheets of Cu (anode) become this, while pure sheets of pure Cu (cathode) become thick.
The reactions in the cell are
At Anode (oxidation)
Cu EMBED Equation.DSMT4 Cu2+ + 2e-
At Cathode (reduction)
Cu2+ + 2e- EMBED Equation.DSMT4 Cu
Thus, there is no net reaction in the cell. However, the net result is the purification of Cu.
(h) S.H.E. acts as anode when connected with the Cu electrode but as cathode with Zn electrode.
See Section 10.3.2
NUMERICAL PROBLEMS (Exercise)
Q14. (c) Calculate the oxidation number of Chromium in the following.
CrCl3 K2CrO4
Oxidation number of Cl=-1 Oxidation number of K = + 1
Oxidation number of Cr=x Oxidation number of Cr= - 2
Oxidation number of Cr Oxidation number of Cr=x
can be calculated as Thus
For CrCl3 For K2CrO4
x + 3( - 1) =0 2(+1) + x +4(-2)=0
x – 3 =0 x – 6 =0
or x =+3 x =+6
K2Cr2O7 CrO3
Oxidation number of K=+1
Oxidation number of O= -2 Oxidation number of O= -2
Oxidation number of Cr=x
Thus Oxidation number of Cr=x
Thus
For K2Cr2O7 For CrO3
2(+1)+2x + 7(-2) =0 x +3(-2)= 0
2x – 12 =0 x – 6 =0
Or x=+12/2 =+6 Or x =+6
Cr2O3 Cr2O72-
Oxidation number of O =-2 Oxidation number of O =-2
Oxidation number of Cr = x Oxidation number of Cr = x
Thus Thus
For Or2O3 For Cr2O72-
2x+3(-2)=0 2x +7(-2) = -2
2x – 6 = 0 2x – 14= - 2 +14
Or x= +6/2 =+3 Or x = + 12/2 =+6
Cr2(SO4)3
Oxidation number of S = +16
Oxidation number of O = -2
Oxidation number of Cr = x
Thus
For Cr2(SO4)3
2x +3[(+6)+4(-2)]=0
2x – 6 =0
Or x=6/2 =+3
(d) Calculate the oxidation numbers of the elements underlined in the following compounds
Na3 PO4 Na2CO3
Oxidation number of Na = +1
Oxidation number of O = -2 Oxidation number of Na = +1 Thus
Oxidation number of P = x Oxidation number of C = x
Thus
For Na3 PO4 For Na2CO3
3(+1) + x + 4(-2) =0 2(+1) + x + 3 (-2) = 0
x – 5 =0 x – 4 =0
Or x = +5 Or x =+4
Cr2(SO4)3 Ca(ClO3)2
Oxidation number of Cr = +3 Oxidation number of Ca = +
Oxidation number of O = - 2 Oxidation number of O = - 2
Oxidation number of S = x Oxidation number of S = x
Thus Thus
For Cr2(SO4)3 For Ca(ClO3)2
2(+3) +3 [ (x)+4(-2)] =0 (+2) +2 [ (x)+3(-2)] =0
3x – 18 =0 2 + 2x -12 =0
Or x = 18/3 = +6 Or x=10/2 = +5
K2 MnO4 HNO3
Oxidation number of K = +1 Oxidation number of H = +1
Oxidation number of O = -2 Oxidation number of O = -2
Oxidation number of Mn = x Oxidation number of N = x
Thus Thus
For K2 MnO4 For HNO3
2(+1) + x + 3(-2) = 0 (+1) + x +3(-2) =0
x – 6 =0 x – 5 =0
Or x = +6 Or x = +15
HPOs
Oxidation number of H = +1
Oxidation number of O = -2
Oxidation number of P = x
Thus
For HPOs
(+1) + x + 3(-2) =0
x - 5 =0
Or x = +5
Try Yourself
H 2O2 , Ca(OCl)2 , NalO2 , Zn(OH)2 , H3PO4,
Q15. (b) Balance the following equations by oxidation number method.
PROBLEM
HNO3 + Hl EMBED Equation.DSMT4 NO + l2 + H2O
Identify the elements, which undergo a change in oxidation number and write their oxidation numbers over the symbols.
+1+5 2(-2) +1 -1 +2 _-2 o
HNO3 + Hl EMBED Equation.DSMT4 NO + l2 + H2O
Determine the no. of electrons gained and lost and equate them.
gain of 3e- x 1 =3 e- (reduction )
+5 -1 +2 o
HNO3 + Hl EMBED Equation.DSMT4 NO + l2 + H2O
lose of l e- x 3 = 3 e- (oxidation)
Balance loss and gain of electrons by multiplying Hl by 3.
HNO3 + 3Hl EMBED Equation.DSMT4 NO + l2 + H2O
Balance the rest of equation by inspection method.
2HNO3 + 6Hl EMBED Equation.DSMT4 2NO + 3l2 + 4H2O
PROBLEM
Br2 + NaOH EMBED Equation.DSMT4 NaBrO3 + NaBr + H2O
Identify the elements which undergo a change in oxidation number and write their oxidation number over the symbols.
o +1 -2 + 1 +1 +5 3(-2) +1 -1
Br2 + NaOH EMBED Equation.DSMT4 NaBrO3 + NaBr + H2O
Since Br2 is involved both in oxidation and reduction, therefore, we shall write the Br=twice. Then determine the no. of electrons gained and lost and equate them.
gain of 2(l) e- x 5 = 10e- (reduction )
o +1 +5 3(-2) +1 -1
Br2 + Br2 + NaOH EMBED Equation.DSMT4 NaBrO3 + NaBr + H2O
loss of 2(5) e- x 1 = 10 e- (reduction)
Balance loss and gain of electrons by multiplying Br2 by 5, in which oxidation occurs.
5Br2 + Br2 + NaOH EMBED Equation.DSMT4 NaBrO3 + NaBr +H2O
Balance the rest of equation by inspection method.
5Br2 + Br2 +12 NaOH EMBED Equation.DSMT4 2NaBrO3 + 10NaBr + 6H2O
Or 6Br2 +12 NaOH EMBED Equation.DSMT4 2NaBrO3 + 10NaBr + 6H2O
Or 3Br2 +6 NaOH EMBED Equation.DSMT4 NaBrO3 + 5NaBr + 3H2O
PROBLEM
Zn + HNO3 EMBED Equation.DSMT4 Zn(NO3) + NO +H2O
Identify the elements, which undergo a change in oxidation number and write their oxidation number over the symbols.
0 +1+5 3(-2) +2+5 +2 -2
Zn + HNO3 EMBED Equation.DSMT4 Zn(NO3) + NO +H2O
In this eq. N is reduced from +5 in HNO3 to +2 in NO. But it’s oxidation state is not change in Cu(NO3)2. therefore, write HNO3 twice and determine the number of electrons gained and lost and equate them.
gain of 3 e- x 2 = 6 e- (reduction )
0 +5 +2 +5 +2 -2
HNO3 +Zn + HNO3 EMBED Equation.DSMT4 Zn(NO3) + NO +H2O
loss of 2 e- x 3 = 6 e- (oxidation )
Use the multiplier obtained above to balance loss and gain of electrons. Thus Cu is multiplied by 3 and HNO3 by 2.
HNO3 +3Zn + 2HNO3 EMBED Equation.DSMT4 Zn (NO3) + NO +H2O
Balance the rest of equation by inspection method.
6HNO3 +3Zn + 2HNO3 EMBED Equation.DSMT4 3Zn(NO3) + 2NO +4H2O
Or
3Zn + 8HNO3 EMBED Equation.DSMT4 3Zn(NO3) + 2NO +4H2O
PROBLEM
+4 2(-2) + -1 +2 2(-1) 0
MnO2 HCI EMBED Equation.DSMT4 MnCl2 + H2O + Cl2
Identify eq. Cl is reduced from – 1 in HCl to 0 in Cl2. But its oxidation state is not changed in MnCl2. therefore, write HCl twice and determine the no. of electrons gained and lost and equate them.
gain of 2 e- x 1 = 2 e- (reduction )
+4 -1 +2 2(-1) 0
HCI+MnO2 HCI EMBED Equation.DSMT4 MnCl2 + H2O + Cl2
loss of 1 e- x 2 = 2 e- (oxidation )
Use the multiplier obtained above to balance loss and gain of electrons.
HCI+MnO2 2HCI EMBED Equation.DSMT4 MnCl2 + H2O + Cl2
Balance the rest of equation by inspection method.
2HCI+MnO2 2HCI EMBED Equation.DSMT4 MnCl2 + 2H2O + Cl2
Or
MnO2 + 4HCI EMBED Equation.DSMT4 MnCl2 + 2H2O + Cl2
PROBLEM
FeSO4 + K2Cr2O7 + H2SO4 EMBED Equation.DSMT4 Fe(SO4) + Cr2(SO4)3 + K2SO4 + H2O
Identify the elements, which undergo a change in oxidation number and write their oxidation numbers over the symbols.
+2 +6 +3 +3
FeSO4 + K2Cr2O7 + H2SO4 EMBED Equation.DSMT4 Fe(SO4) + Cr2(SO4)3 + K2SO4 + H2O
Determine the no of electrons gained and lost and equate them.
gain of 2(3) e- x 1 = 6 e- (reduction )
+2 +6 +3 +3
FeSO4 + K2Cr2O7 + H2SO4 EMBED Equation.DSMT4 Fe(SO4) + Cr2(SO4)3 + K2SO4 + H2O
loss of 1(2) e- x 3 = 6 e- (oxidation )
Use the multiplier obtained above to balance loss and gain of electrons
6FeSO4 + K2Cr2O7 + H2SO4 EMBED Equation.DSMT4 3Fe (SO4) + Cr2 (SO4)3 + K2SO4 + H2O
Balance the rest of equation by inspection method.
6FeSO4 + K2Cr2O7 + 7H2SO4 EMBED Equation.DSMT4 3Fe(SO4) + Cr2(SO4)3 + K2SO4 + 7H2O
PROBLEM
Cu + H2SO4 CuSO4 + SO2 + H2O
Identify the elements, which undergo a change in oxidation number and write their oxidation number over the symbols.
0 +6 +2 +4 2((-2)
Cu + H2SO4 EMBED Equation.DSMT4 CuSO4 + SO2 + H2O
In this equation S is reduced from +6 in H2SO4 to +4 in SO2. But it’s oxidation state is not changed in CuSO4. Therefore, write H2SO4 twice and determine the no. of electrons gained and lost equate them.
gain of 2 e- x 1 = 2 e- (reduction )
0 +6 +2 +4
H2SO4+ Cu + H2SO4 EMBED Equation.DSMT4 CuSO4 + SO2 + H2O
loss of 2 e- x 1 = 2 e- (oxidation )
Use the multiplier obtained above to balance loss and gain of electrons.
H2SO4+ Cu + H2SO4 EMBED Equation.DSMT4 CuSO4 + SO2 + H2O
Balance the rest of equation by inspection method.
H2SO4+ Cu + H2SO4 EMBED Equation.DSMT4 CuSO4 + SO2 + 2H2O
Or
Cu + 2H2SO4 EMBED Equation.DSMT4 CuSO4 + SO2 + 2H2O
PROBLEM
H2SO4+ Hl EMBED Equation.DSMT4 SO4 + l2 + H2O
Identify the elements, which undergo a change in oxidation number and wirte their oxidation number over the symbols.
+6 +1 -1 +4 2(-2) 0
H2SO4+ Hl EMBED Equation.DSMT4 SO4 + l2 + H2O
Determine the no. of electrons gained and lost and equate them.
gain of 2 e- x 1 = 2 e- (reduction )
H2SO4+ Hl EMBED Equation.DSMT4 SO4 + l2 + H2O
loss of 2 e- x 1 = 2 e- (oxidation )
Use the multiplier obtained above to balance loss and gain of electrons
H2SO4+ 2Hl EMBED Equation.DSMT4 SO4 + l2 + H2O
Balance the rest of equation by inspection method.
H2SO4+ 2Hl EMBED Equation.DSMT4 SO4 + l2 + 2H2O
PROBLEM
NaCl + MnO2 + H2SO4 EMBED Equation.DSMT4 Na2SO4 + MnSO4 + H2O + Cl2
Identify the elements, which undergo a change in oxidation number and write their oxidation number over the symbols.
+1 -1 +4 2(-2) +2 0
NaCl + MnO2 + H2SO4 EMBED Equation.DSMT4 Na2SO4 + MnSO4 + H2O + Cl2
Determine the number of electrons gained and lost and equate them.
gain of 2 e- x 1 = 2 e- (reduction )
-1 +4 +2 0
NaCl + MnO2 + H2SO4 EMBED Equation.DSMT4 Na2SO4 + MnSO4 + H2O + Cl2
loss of 2 e- x 1 = 2 e- (oxidation )
Use the multiplier obtained above to balance loss and gain of electrons
2NaCl + MnO2 + H2SO4 EMBED Equation.DSMT4 Na2SO4 + MnSO4 + H2O + Cl2
Balance the rest of equation by inspection method.
2NaCl + MnO2 + H2SO4 EMBED Equation.DSMT4 Na2SO4 + MnSO4 + H2O + Cl2
PROBLEM
K2Cr2O7 + HCl EMBED Equation.DSMT4 KCl + CrCl3 + H2O + Cl2
Identify the elements, which undergo a change in oxidation number and write their oxidation number over the symbols.
+6 - 1 +3 0
K2Cr2O7 + HCl EMBED Equation.DSMT4 KCl + CrCl3 + H2O + Cl2
In this eq. Cl is oxidation from -1 in HCl to 0 in Cl2. But it’s oxidation state is not change in KCl / or CrCl3. therefore, write HCl twice and determine the no. of electrons gained and lost and equate them.
gain of 2(3) =6 e- x 1 = 3 e- (reduction )
+6 - 1 +3 0
HCl+ K2Cr2O7 + HCl EMBED Equation.DSMT4 KCl + CrCl3 + H2O + Cl2
loss of 1 e- x 6 = 6 e- (oxidation )
Use the multiplier obtained above to balance loss and gain of electrons
HCl+ K2Cr2O7 + 6HCl EMBED Equation.DSMT4 KCl + CrCl3 + H2O + Cl2
Balance the rest of equation by inspection method.
8HCl+ K2Cr2O7 + 6HCl EMBED Equation.DSMT4 2KCl +2CrCl3 + 7H2O + 3Cl2
Or
K2Cr2O7 + 14HCl EMBED Equation.DSMT4 2KCl + 2CrCl3 + 7H2O + 3Cl2
Q16. (b) Balance the following equation by ion-election method
PROBLEM
Sn2+ + Fe3+ EMBED Equation.DSMT4 Sn4+ + Fe2+
Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reaction.
Fe3+ EMBED Equation.DSMT4 Fe2+
Sn2+ EMBED Equation.DSMT4 Sn4+
Write down the number of electrons gained and lost in each half reaction
Fe3+ +le- EMBED Equation.DSMT4 Fe2+ (reduction half reaction) ______(1)
Sn2+ EMBED Equation.DSMT4 Sn4+ + 2e- (oxidation half reaction) ______(2)
Equate the total number of electrons gained and lost by multiplying eq. (1) by 2, and then add them.
2Fe3+ +2e- EMBED Equation.DSMT4 2Fe2+
Sn2+ EMBED Equation.DSMT4 Sn4+ + 2e-
Sn2+ +2Fe3+ EMBED Equation.DSMT4 Sn4+ +2Fe2+
PROBLEM
H+ + Cl- + Cr2O72- EMBED Equation.DSMT4 Cr3++ Cl2
Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions.
Cr2O72- EMBED Equation.DSMT4 Cr3+ (reduction half reaction)
2Cl- EMBED Equation.DSMT4 Cl2 (oxidation half reaction)
Balance oxygen by adding H2O.
Cr2O72- EMBED Equation.DSMT4 Cr3+ + 7H2O
2Cl- EMBED Equation.DSMT4 Cl2
Balance hydrogen by adding H+ ions.
14H+ + Cr2O72- EMBED Equation.DSMT4 2Cr3+ + 7H2O
2Cl- EMBED Equation.DSMT4 Cl2
Write down the number of electrons gained and lost in each half reaction.
14H+ + Cr2O72- + 6e- EMBED Equation.DSMT4 2Cr3+ + 7H2O _________(1)
2Cl- EMBED Equation.DSMT4 Cl2 +2e- _________(2)
Equate the total number of electrons gained and lost by multiplying eq.(2) by 3. And then add the two half reactions.
14H+ + Cr2O72- + 6e- EMBED Equation.DSMT4 2Cr3+ + 7H2O
6Cl- EMBED Equation.DSMT4 3Cl2 + 6e-
Cr2O72- +14H+ + 6e- EMBED Equation.DSMT4 2Cr3+ +NO2+ 7H2O
PROBLEM (Acidic Media)
Cu + NO3-1 + H+ EMBED Equation.DSMT4 Cu+2 + NO2+ + H2O
Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions.
NO3-1 EMBED Equation.DSMT4 NO2 (reduction half reaction)
Cu EMBED Equation.DSMT4 Cu2+ (oxidation half reaction)
Balance oxygen by adding H2O.
NO3-1 EMBED Equation.DSMT4 NO2 +H2O
Cu EMBED Equation.DSMT4 Cu2+
Balance hydrogen by adding H+ ions.
2H+ + NO3-1 EMBED Equation.DSMT4 NO2
Cu EMBED Equation.DSMT4 Cu2+
Write down the number of electrons gained and lost in each half reaction
4H+ + NO3-1 + le- EMBED Equation.DSMT4 NO2 +H2O __________(1)
Cu EMBED Equation.DSMT4 Cu2+ + 2e- __________(2)
Equate the total number of electrons gained and lost by multiplying eq. (1) by 2. And then add the two half reactions.
4H+ + 2NO3-1 + 2e- EMBED Equation.DSMT4 2NO2 +2H2O
Cu EMBED Equation.DSMT4 Cu2+ + 2e-
2NO3-1 +4H+ + Cu EMBED Equation.DSMT4 Cu2+ + 2NO2 +2H2O
PROBLEM (Acidic Media)
Cr2O72- + H3AsO3 EMBED Equation.DSMT4 Cr3 + H3AsO4
Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions.
Cr2O72- EMBED Equation.DSMT4 2Cr3+ (reduction half reaction)
H3AsO3 EMBED Equation.DSMT4 H3AsO4 (oxidation half reaction )
Balance oxygen by adding H2O.
Cr2O72- EMBED Equation.DSMT4 2Cr3 + 7H3O
H2O + H3AsO3 EMBED Equation.DSMT4 H3AsO4
Balance hydrogen by adding H+ ions.
14H+ +6e-+ Cr2O72- + EMBED Equation.DSMT4 2Cr3 + 7H3O ________(1)
H2O +3 H3AsO3 EMBED Equation.DSMT4 3 H3AsO3 + 2e- + 2H+ ________(2)
Equate th total number of electrons gained and lost by multiplying eq. (2) by 3. And then add the two half reactions.
14H+ +6e-+ Cr2O72- + EMBED Equation.DSMT4 2Cr3 + 7H3O
3H2O +3 H3AsO3 EMBED Equation.DSMT4 3 H3AsO3 + 6e- + H+
Cr2O72- +8H+ + 3H3AsO3 EMBED Equation.DSMT4 2Cr3 + 3H3AsO4 + 4H2O
PROBLEM (Acidic Media)
MnO4-+ Cr2O72- EMBED Equation.DSMT4 Mn2+ + CO2
Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions.
MnO4- EMBED Equation.DSMT4 Mn2+ (reduction half reaction)
Cr2O72- EMBED Equation.DSMT4 2CO2 (oxidation half reaction)
Balance oxygen by adding H2O.
MnO4- EMBED Equation.DSMT4 Mn2+ + 4H2O
Cr2O72- EMBED Equation.DSMT4 2CO2
Balance hydrogen by adding H+ ions.
8H+ +MnO4- EMBED Equation.DSMT4 Mn2+ + 4H2O
Cr2O72- EMBED Equation.DSMT4 2CO2
Write down th number of electrons gained and lost in each half reaction
5e-+ 8H+ +MnO4- EMBED Equation.DSMT4 Mn2+ + 4H2O __________(1)
Cr2O72- EMBED Equation.DSMT4 2CO2 +2e-1 __________(2)
Equate th total number of electrons gained and lost by multiplying eq. (2) by 5 and eq. (1) by 2. And then add the two half reactions.
10e-+ 16H+ +2MnO4- EMBED Equation.DSMT4 2Mn2+ + 8H2O + 10CO2
5Cr2O72- EMBED Equation.DSMT4 10CO2 +10e-1
5C2O42-+ 16H+ +2MnO4- EMBED Equation.DSMT4 2Mn2+ + 8H2O + 10CO2
PROBLEM (Acidic Media)
Fe2+ + C2O72- EMBED Equation.DSMT4 Cr3+ + Fe3+
Identify the elements, which undergo oxidation and reduction and split up the reactions into oxidation and reduction half reactions.
C2O72- EMBED Equation.DSMT4 Cr3+ (reduction half reaction)
Fe2+ EMBED Equation.DSMT4 Fe3+ (oxidation half reaction)
Balance oxygen by adding H2O.
C2O72- EMBED Equation.DSMT4 2Cr3+ + 7H2O.
Balance hydrogen by adding H+ ions.
14H++ C2O72- EMBED Equation.DSMT4 2Cr3+ + 7H2O.
Fe2+ EMBED Equation.DSMT4 Fe3+
Write down th number of electrons gained and lost in each half reaction
14H+ +6e-+ Cr2O72- EMBED Equation.DSMT4 2Cr3 + 7H3O ________(1)
Fe2+ EMBED Equation.DSMT4 Fe3+ + le- ________(2)
Equate the total number of electrons gained and lost by multiplying eq. (2) by 6. And then add the two half reactions.
14H+ +6e-+ Cr2O72- EMBED Equation.DSMT4 2Cr3 + 7H3O
6Fe2+ EMBED Equation.DSMT4 6Fe3+ + 6e-
Cr2O72- +14H+ +6e- EMBED Equation.DSMT4 6Fe2++ 2Cr3 + 7H3O
PROBLEM (Acidic Media)
lO3- + AsO33- EMBED Equation.DSMT4 l- + AsO43-
Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction and reduction half reactions.
lO3- EMBED Equation.DSMT4 l- (reduction half reaction)
AsO33- EMBED Equation.DSMT4 AsO43- (oxidation half reaction)
Balance oxygen by adding H2O.
lO3- EMBED Equation.DSMT4 l- + 3H2O
H2O+ AsO33- EMBED Equation.DSMT4 AsO43-
Balance oxygen adding H+ ions.
6H+ + lO3- EMBED Equation.DSMT4 l- + 3H2O
H2O+ AsO33- EMBED Equation.DSMT4 AsO43- + 2H+
Write down the number of electrons gained and lost in each half reaction
6e- + 6H+ + lO3- EMBED Equation.DSMT4 l- + 3H2O ___________(1)
H2O+ AsO33- EMBED Equation.DSMT4 AsO43- + 2H+ + 2e - ___________(2)
Equate the total number of electrons gained and lost by multiplying eq. (2) by 3. And then add the two half reactions.
6e- + 6H+ + lO3- EMBED Equation.DSMT4 l- + 3H2O
3H2O+ 3AsO33- EMBED Equation.DSMT4 3AsO43- + 6H+ + 6e -
lO3- + 3AsO33 EMBED Equation.DSMT4 3AsO43- +l-
PROBLEM (Acidic Media)
Cr3+ + BiO3 EMBED Equation.DSMT4 Cr2O72- + Bi3+
Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions.
BiO3 EMBED Equation.DSMT4 Bi3+ (reduction half reaction)
Cr3+ EMBED Equation.DSMT4 Cr2O72- (oxidation half reaction)
Balance oxygen by adding H2O.
BiO3 EMBED Equation.DSMT4 Bi3+ + 3H2O
7H2O + Cr3+ EMBED Equation.DSMT4 Cr2O72-
Balance hydrogen by adding H+ ions.
6H+ + BiO3 EMBED Equation.DSMT4 Bi3+ + 3H2O
7H2O + Cr3+ EMBED Equation.DSMT4 Cr2O72- + 14H+
Write down the number of electrons gained and lost in each half reaction
2e- + 6H+ + BiO3 EMBED Equation.DSMT4 Bi3+ + 3H2O ________(1)
7H2O + 2Cr3+ EMBED Equation.DSMT4 Cr2O72- + 14H+ __ _______(2)
Equate the total number of electrons gained and lost by multiplying eq.(1) by 3. And then add the tow half reactions.
6e- + 8H+ + 3BiO3 EMBED Equation.DSMT4 3Bi3+ + 9H2O
7H2O + 2Cr3+ EMBED Equation.DSMT4 Cr2O72- + 14H+
3BiO3 + 2Cr3+ EMBED Equation.DSMT4 3Bi3+ + Cr2O72- + 2H2O
PROBLEM (Acidic Media)
OCl- + S2O32- EMBED Equation.DSMT4 Cl- + S4O62-
Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions.
OCl- EMBED Equation.DSMT4 Cl- (reduction half reaction)
S2O32- EMBED Equation.DSMT4 S4O62- (oxidation half reaction)
Balance oxygen by adding H2O.
OCl- EMBED Equation.DSMT4 Cl- + H2O
2 S2O32- EMBED Equation.DSMT4 S4O62-
Balance hydrogen by adding H+ ions.
OCl- + 2H+ EMBED Equation.DSMT4 Cl- + H2O
2 S2O32- EMBED Equation.DSMT4 S4O62-
Write down the number of electrons gained and lost in each half reaction
2e- + OCl- + 2H+ EMBED Equation.DSMT4 Cl- + H2O ________(1)
2 S2O32- EMBED Equation.DSMT4 S4O62-+ 2e- ________(2)
Add the two half reactions.
2e- + OCl- + 2H+ EMBED Equation.DSMT4 Cl- + H2O
2 S2O32- EMBED Equation.DSMT4 S4O62-+ 2e-
2 S2O32- + OCl- + 2H+ EMBED Equation.DSMT4 S4O62-+ CO2
PROBLEM (Acidic Media)
MnO4- + C2O42 EMBED Equation.DSMT4 MnO2 + CO2
Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions.
MnO4- EMBED Equation.DSMT4 MnO2 (reduction half reaction)
C2O42 EMBED Equation.DSMT4 CO2 (oxidation half reaction)
Add two OH- ions for one each oxygen atom on appropriate side.
MnO4- EMBED Equation.DSMT4 MnO2 + 4 OH-
C2O42 EMBED Equation.DSMT4 CO2
Balance hydrogen by adding H2O ions.
2H2O + MnO4- EMBED Equation.DSMT4 MnO2 + 4 OH-
C2O42 EMBED Equation.DSMT4 CO2
Write down the number of electron gained and lost in each half reaction
3e-+ 2H2O + MnO4- EMBED Equation.DSMT4 MnO2 + 4 OH- ________(1)
C2O42 EMBED Equation.DSMT4 CO2 + 2e- ________(2)
Equate the total number of electrons gained and lost by multiplying eq. (1) by 2 and eq. (2) by 3. And then add the two half reactions.
6e-+ 4H2O + 2MnO4- EMBED Equation.DSMT4 2MnO2 + 8OH-
3C2O42 EMBED Equation.DSMT4 6CO2 + 6e-
3C2O42-+2MnO4- + 4H2O EMBED Equation.DSMT4 2MnO2 + 8OH-
PROBLEM (Acidic Media)
MnO4- + CN- EMBED Equation.DSMT4 MnO2 + CNO-
Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions.
MnO4- EMBED Equation.DSMT4 MnO2 (reduction half reaction)
CN- EMBED Equation.DSMT4 CNO- (oxidation half reaction)
Balance hydrogen and oxygen by adding H2O and OH- ions.
2H2O +MnO4- EMBED Equation.DSMT4 MnO2 +4 OH- _________(1)
CN- +2 OH- EMBED Equation.DSMT4 CNO- + H2O _________(2)
Write down the number of electrons gained and lost in each half reaction
3e-+ 2H2O + MnO4- EMBED Equation.DSMT4 MnO2 + 4OH- __________(1)
CN- +2OH- EMBED Equation.DSMT4 CNO- + H2O +2e- _________ __(2)
Equate the total number of electrons gained and lost by multiplying eq.(1) by 3. And then add the tow half reactions.
6e-+ 4H2O + 2MnO4- EMBED Equation.DSMT4 2MnO2 + 8OH- ___________(1)
3CN- + 6OH- EMBED Equation.DSMT4 3CNO- + 3H2O +6e- _________ __(2)
3CN- + H2O +2MnO4- EMBED Equation.DSMT4 2MnO2+ 6OH-+3CNO-
PROBLEM (Acidic Media)
H3AsOs + Cr2O72- EMBED Equation.DSMT4 Cr3+ + H3AsO4
Identify the elements, which undergo oxidation and reduction and split up the reaction into oxidation and reduction half reactions.
Cr2O72- EMBED Equation.DSMT4 Cr3+ (reduction half reaction)
H3AsOs EMBED Equation.DSMT4 H3AsO4 (oxidation half reaction)
Balance oxygen by adding H2O.
Cr2O72- EMBED Equation.DSMT4 Cr3+ +7 H2O
H3AsOs + H2O EMBED Equation.DSMT4 H3AsO4
Balance hydrogen by adding H+ ions.
14H+ +Cr2O72- EMBED Equation.DSMT4 Cr3+ +7 H2O
H3AsOs + H2O EMBED Equation.DSMT4 H3AsO4 + 2H+
Write down the number of electrons gained and lost in each half reaction.
14H+ + 6e- +Cr2O72- EMBED Equation.DSMT4 Cr3+ +7 H2O _______(1)
H3AsOs + H2O EMBED Equation.DSMT4 H3AsO4 + 2H+ + 2e- _______(2)
Equate the total number of electrons gained and lost by multiplying eq.(1) by 3. And then add the tow half reactions.
14H+ +6e- +Cr2O72- EMBED Equation.DSMT4 2Cr3+ +7 H2O ________(1)
3H3AsOs + 3H2O EMBED Equation.DSMT4 3H3AsO4 + 6H+ + 6e- ________(2)
Cr2O72- +8H+ + 3H3AsOs EMBED Equation.DSMT4 3H3As
3 comments:
yeh jo likha hota hai see section 10.1 ...yeh section kahan hai???
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10th class past papers
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